3 red, 2 white and 3 green balls are available in a bag. 2 balls are drawn one by one from the bag.

Surit Datta   9 years AGO

total number of balls are 3 + 2 +3 = 8
i have to select the balls such that,
one is white
another is green
so probability is
2C1*3C1/8C2
which is 6/28 = 3/14
so option 2
that's it.

Like? Yes (19) | No (3)  

Supriya Mondal   9 years AGO

Probability will be=(2C1*3C1)/8C2
=3/14
So,option 2)3/14 is the right answer

Like? Yes (4) | No (1)  

Yathiraj   9 years AGO

Total Number of balls = 3+2+3 = 8
Let S be the sample space.
Then n(S) = Number of ways of drawing 2 balls out of 8
=> 8C2 => 28

There are 6 favorable combinations of getting one White and one Green ball. (3 green and 2 white balls)
(G1 W1), (G1 W2),
(G2 W1), (G2 W2),
(G3 W1), (G3 W2)

Hence, Probability = 6/28 => 3/14
Option 2.

Like? Yes (3) | No (1)  

MD SOHAIL AKHTER   9 years AGO

2/8*3/7*2=3/14

Like? Yes (1) | No   

Pranati Kundu   9 years AGO

1st ball white and another ball green =2/8*3/7 or 1st ball green and another ball white =3/8*2/7
prob= 2/8*3/7 + 3/8*2/7
=3/14

Like? Yes (1) | No   

sindhura   9 years AGO

(2c1*3c1)/8c2=6/28=3/14

Like? Yes (2) | No (1)  

sourav pal   9 years AGO

Let S be the sample Space
Then,n(S)=number of ways of drawing 2 balls out of 8.
=8C2=28
Let E be the event of drawing 1 red and 1 green
so,n(E)=2C1*3C1=6
P(E)=n(E)/n(S)=6/28=3/14

Like? Yes (1) | No   

Preethi   9 years AGO

prob to get 1 green out of 8 is 3/8
prob to get 1 white among remaining 7 is 2/7
so 2*2*3/8*7
= 3/14

Like? Yes (1) | No   

Sunil Chavan   9 years AGO

(2C1*3C1)/(8C2)

Like? Yes (1) | No   

Srinivas   9 years AGO

(2c1 * 3c1)/8c2

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nitin kishore   9 years AGO

There can be two possible ways:
a) first ball green and second ball white
b) first ball white and second ball green

so probability:-
(3/8)*(2/7)+(2/8)*(3/7)
3/14

Like? Yes (1) | No   

madhusudhana   9 years AGO

total number of combinations is 28
in that number of favorable combination (1 white and 1 green) will be 6
So the probability of 1 white and 1 green will be 6/28 I.,e 3/14

Like? Yes (1) | No   

Pankaj kumar   9 years AGO

probability of getting one white ball=2.
probability of getting one green ball=3.
total probability=28.
p(w)=2c1=2
p(G)=3c1=3
p(T)=(p(w)*p(G))/(p(T))=3/14.

Like? Yes (1) | No   

vasavi kandagatla   9 years AGO

(2c1*3c1)/(8c2)=6/28=3/14

Like? Yes (1) | No   

Rahul pal   9 years AGO

2C1×3c1/8c2

Like? Yes (1) | No   

neeraj   9 years AGO

(2 white me se 1)*(3 green me se 1)
/(total 8 me se 2 ball)

Like? Yes (1) | No   

Divya Jain   9 years AGO

(2C1*3C1)/8C2=3/14

Like? Yes (1) | No   

aaditya raj   9 years AGO

if first ball is green then probablity is 3/8 and second ball is white probablity is 2/7 if we chose first white then probablity is 2/8 and probablity of green is 3/7 then total probality is = (3/8)*(2/7)+(2/8)*(3/7) = 3/14

Like? Yes (1) | No   

nikhil   9 years AGO

white green = 2/8 * 3/7
green white = 3/8 * 2/7

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Josepheena Joseph   9 years AGO

(2C1x3C1)/8C2 = 3/14

Like? Yes (1) | No   

vishwa   9 years AGO

3/14

Like? Yes (1) | No   

Mounis Swaleh   9 years AGO

(2c1 x 3c1) / 8c2 = 6/ 28 = 3/14

Like? Yes (1) | No   

rayvadera meet    9 years AGO

Probability= (2 into 3/8 into 7) into 2

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karthiga   9 years AGO

(2C1 *3C1)/ 8C2;
= 2*3/(8*7*6!)/6!*2!
= 3/14.
Answer is 3/14

Like? Yes (1) | No   

SAHANA M   9 years AGO

(2c1*3c1)/8c2

Like? Yes (1) | No   

Arpitha Loka   9 years AGO

total balls=8
2 are drawn from them so 8c2 ways=28
drawing one ball from white and other from green=2c1x3c1
=6 ways
probability of one ball having white and other green =6/28
i.e.,3/14.

Like? Yes (1) | No   

Shintala Divya   9 years AGO

(2c1*3c1)/8c2=3/14

Like? Yes (1) | No   

sudhakar   9 years AGO

option 2)3/14

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santhoshkumar pariselli   9 years AGO

(3c1*2c1)/(2+3+3)c2=6/28=3/14

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jaganmohan rao .pola   9 years AGO

3/14

Like? Yes (1) | No   

veena pavani   9 years AGO

one ball is white and green so 2c1*3c1=6
two balls drawn from the bag so have to consider total no of balls 8c2=28
probability=6/28=3/14

Like? Yes (1) | No   

ramakrishna   9 years AGO

Let S be the sample space.

Then, n(S) = number of ways of drawing 2 balls out of 8
= 8C2
n(S)=(8*7)/2*1=28


Let E=event of getting one whiteand other green
=2c1*3c1=2*3
n(E)=6

p(E)=n(E)/n(S)
=6/28
=3/14

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pooja   9 years AGO

(2c1*3c1)/8c2=3/14

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ABHINAV PUSHPAM   9 years AGO

Option 2

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rohit Prakash gupta   9 years AGO

(2C1*3C1)/8C1=(2*3)/(4*7)=3/14

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adrino   9 years AGO

[(2/8)*(3/7]+[(3/8)*(2/7)]
3/14

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Jay parmar   9 years AGO

option 2 is correct

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RAKESH   9 years AGO

2/8 * 3/7 + 3/8 *2/7 = 3/14

Like? Yes (1) | No (1)  

aman aditya   9 years AGO

sample space=8c2=28
favorable outcome=2c1 *3c1=6
so
probability=6/28=3/14

Like? Yes (1) | No (1)  

tasho k   9 years AGO

Ball is drawn one by one..
no mention of replacement.
so
2c1×3c1/8c2=3/14

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Ajinkya Kulkarni   9 years AGO

3/14

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KARRI V V S B VAMSI KRISHNA REDDY   9 years AGO

total no. of balls = 3+2+3 = 8
2 balls should be selected from 8 balls = 8c2 = 28
1 ball from white and 1 ball from green = 2c1 * 3c1 = 6
Probability = 6/28 = 3/14

Like? Yes (1) | No (1)  

bollina lakshmi bhargavi   9 years AGO

2c1+3c1/8c1+7c1=5/15=1/3

Like? Yes (1) | No (1)